Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
answer: elements are on the periodic table
Convection occurs because the oceanic waters heat up becoming less dense. .This water moves above the cooler water, and give off its heat to the surrounding environment. As it cools, it begins to sink, and the process begins again.
It depends on what kind of reaction it is. it could be 10 moles too, but it can be less, for example, if the products looked like this
CO2+2 H2O the answer would be:
x mol CO2 - 10 mol H20
1 mol CO2 - 2 mol H2O
Answer:
Explanation:
{\displaystyle {}^{n}x}{}^{n}x, for n = 2, 3, 4, …, showing convergence to the infinitely iterated exponential between the two dots
In mathematics, tetration (or hyper-4) is an operation based on iterated, or repeated, exponentiation. It is the next hyperoperation after exponentiation, but before pentation. The word was coined by Reuben Louis Goodstein from tetra- (four) and iteration.
Under the definition as repeated exponentiation, the notation {\displaystyle {^{n}a}}{\displaystyle {^{n}a}} means {\displaystyle {a^{a^{\cdot ^{\cdot ^{a}}}}}}{\displaystyle {a^{a^{\cdot ^{\cdot ^{a}}}}}}, where n copies of a are iterated via exponentiation, right-to-left, I.e. the application of exponentiation {\displaystyle n-1}n-1 times. n is called the "height" of the function, while a is called the "base," analogous to exponentiation. It would be read as "the nth tetration of a".
Tetration is also defined recursively as
{\displaystyle {^{n}a}:={\begin{cases}1&{\text{if }}n=0\\a^{\left(^{(n-1)}a\right)}&{\text{if }}n>0\end{cases}}}{\displaystyle {^{n}a}:={\begin{cases}1&{\text{if }}n=0\\a^{\left(^{(n-1)}a\right)}&{\text{if }}n>0\end{cases}}},
allowing for attempts to extend tetration to non-natural numbers suc